Cult of Gravity

tan(1) = earth / jupiter
tan(1) jupiter = earth
928,081,020,000 * 0.0174550649 = 16,199,714,437

16,199,714,437 * 2 = 32,399,428,874

source

tan(1) = earth / jupiter
tan(1) jupiter = earth
928,081,020,000 * 0.0174550649 = 16,199,714,437

16,199,714,437 * 2 = 32,399,428,874

source

Apologies but could be a little more thorough, it has been a while since I have done trig.  I am not sure what "earth" means in this equation, same with "jupiter".  Is it the distance to Earth and Jupiter respectively?  From what then?  Thanks for helping me through this.

I am not sure what "earth" means in this equation, same with "jupiter".  Is it the distance to Earth and Jupiter respectively?

earth would be half the distance between the two telescopes (on Earth). jupiter would be the alleged distance from Earth to Jupiter.

Quote from: Rama Set on March 21, 2013, 11:16:08 AM

I am not sure what "earth" means in this equation, same with "jupiter".  Is it the distance to Earth and Jupiter respectively?

earth would be half the distance between the two telescopes (on Earth). jupiter would be the alleged distance from Earth to Jupiter.

I see.  Two comments:

1. I think that website has a typo in how it phrases the distance to Jupiter.  www.universetoday.com says, "At their most distant from each other [the Earth and Jupiter] are 928,081,020 million km apart."  It should just say, "928,081,020 kms apart".  Not a big deal, but it does throw your solution off by 3 orders of magnitude. 

2. Why are you using a point halfway between the two telescopes?  If we call one telescope A, the second B, and Jupiter C, then the length AC is 928,081,020 kms, the lines AB and AC are at 90 degrees to one another, and the lines AC and BC are at 1 degree to each other.  So in the equation you presented you are trying to solve the length of the line AB so we have:

tan 1=0.01745506492
AC=928,081,020 kms
tan 1=AB/AC
Solve for AB

0.01745506492=AB/928,081,020
AB=0.01745506492*928,081,020
AB=16,199,714.46 kms

Do you agree with this?

Muggsy, are you deliberately misunderstanding what I said, or are you just plain stupid? I never mentioned using tan anywhere in my description. Also, I did say "verify that differences in their angles can be measured accurately" when talking about the telescopes. Angles can be measured accurately to less than a second of arc (1/3600th of a degree) without too much difficulty, so an entire degree of separation (which Rama Set kindly calculated accurately) is not necessary.

 FYI, one second of separation is 4,499km, in the context of measuring the distance to Jupiter.

Muggsy, are you deliberately misunderstanding what I said, or are you just plain stupid? I never mentioned using tan anywhere in my description. Also, I did say "verify that differences in their angles can be measured accurately" when talking about the telescopes. Angles can be measured accurately to less than a second of arc (1/3600th of a degree) without too much difficulty, so an entire degree of separation (which Rama Set kindly calculated accurately) is not necessary.

 FYI, one second of separation is 4,499km, in the context of measuring the distance to Jupiter.

I am trying to figure out how to do the transformation of the viewing angle as the telescopes move along the arc of the Earth's surface.  Any help?

No worries Rama, I'll try to give you both FE and RE methods if I can

RE (aka 'the real world'): This will be easier with an equatorial mount, which you can set for your latitude and it will save you the trouble of working out that part of the movement. Then you just need to take into account that the earth rotates one degree every 4 minutes, or one minute of angle (minute of arc, or MOA) every 4 seconds, so you will need to adjust your telescope accordingly. Most of the medium to high end telescopes for amateur use have built-in tracking, which makes this adjustment for you. So, as the earth rotates, the angle increases towards the West.

FE: This depends greatly on which FE model you want to use, but for the two more common ones (flat stationary disk, with the heavens describing a circle overhead, or stationary heavens with a rotating disk), you will need to ensure your telescopes turntable is level, then just turn it clockwise (looking from above) by the same amount as for RE (one degree every 4 minutes, or one MOA every 4 seconds).

I hope that was the answer you were after! If not, let me know, and I'll have another crack at answering you.

For some truly awesome viewing, and loads of information on astronomy, check out this channel: http://www.youtube.com/user/DeepSkyVideos

I am not sure if we mean the same thing, I will restate it just to be extra clear:

Someone looking at Jupiter from the North Pole sees it at 35 degrees up from the horizon. Where would a person at 73 N and the same longitude see Jupiter in terms of elevation from the horizon assuming they were looking at the exact same time?

My apologies Rama Set, I got your question 90 degrees out!

This is actually even simpler to work out (well, it is on a round earth at least!). If a person at the North pole sees Jupiter at 35 degrees above the horizon, then for every 60 nautical miles, or one degree of latitude, that you move South, Jupiter will 'rise' one degree in the sky. To put that in the context of your question, at 73N, Jupiter would appear to be 35 + 90 - 73 = 52 degrees above the horizon (35 is it's apparent elevation at 90 degrees latitude (north pole), and the difference in elevation is simply the difference in latitude (90 - 73 = 17)).

For a FE model, your results would depend greatly on the altitude of the celestial bodies. First you would need to calculate a position fir Jupiter, using whatever altitude it's supposed to be at. I'll write you an equation:

e = elevation relative to the horizon (in degrees)
a = altitude of object to be observed
d = horizontal distance to object (note, this is not the line of sight distance! Rather, it is the distance along the ground, assuming the ground is entirely flat)

d = a / tan(e)

So, for Jupiter having an altitude of 3500 miles (chosen because it seems to suit some FE models), and your angle of 35 degrees: d = 3500 / tan(35) = 4998.5 miles. If we now go to 73N, we have traveled 17 * 60 = 1020 nautical miles. Better convert those to statute, since that's what we're doing our calculations in = 1147.5 miles. So our horizontal distance to Jupiter is now 4998.5 - 1147.5 = 3851 miles. Rearranging our equation to give us the angle:
tan(e) = a / d
e = tan^-1 (a / d)
Plug in the numbers:
e = tan^-1 (3500 / 3851) = 42.3 degrees

Okay, I know that was really long winded, but I hope it answered your question!

Quote from: Roundy the Truthinessist on March 21, 2013, 06:56:21 AM

Quote from: Rama Set on March 21, 2013, 05:17:31 AM

Quote from: Roundy the Truthinessist on March 20, 2013, 10:31:39 PM

Quote from: Rama Set on March 20, 2013, 10:16:03 PM

Quote from: Roundy the Truthinessist on March 20, 2013, 09:49:42 PM

Quote from: Scintific Method on March 20, 2013, 08:26:12 PM

Quote

Here we go again. "The earth is round, just like everything in the sky. Except for the things that aren't."

If you think about that for just a couple of seconds, you'll realise that the things that aren't round are all quite a bit smaller than the things that are round

This is incorrect.  I think you'll find that galaxies, according to standard RE theory, are quite large.

The large things that are in galaxies are round, the small ones less so.

I'm not talking about the things in galaxies.  I'm talking about galaxies.  They are huge and they are flat.

Quote

The gravity in a galaxy locally is very weak unlike inside a star.

Your point?  On the whole the gravity in a galaxy is strong enough to hold together billions of stars, or so we are told.  That doesn't seem very weak to me.

And it pulls them together into a flat shape.

The average force of gravity is weaker than in a star or planet. If it were stronger than it would collapse to a sphere. The competing forces, solar winds, gravity from other galaxies, prevents this.

So everything in the universe is compelled by gravity to pull into a sphere... unless it's too small... or unless it's too large... got it.  In other words,

Quote

"The earth is round, just like everything in the sky. Except for the things that aren't."

Not what I said, but take your meaning as you wish.

No, I'd rather we both understand this properly.  How does what I said differ from what you said?

Quote from: Rama Set on March 21, 2013, 07:23:38 AM

Quote from: Roundy the Truthinessist on March 21, 2013, 06:56:21 AM

Quote from: Rama Set on March 21, 2013, 05:17:31 AM

Quote from: Roundy the Truthinessist on March 20, 2013, 10:31:39 PM

Quote from: Rama Set on March 20, 2013, 10:16:03 PM

Quote from: Roundy the Truthinessist on March 20, 2013, 09:49:42 PM

Quote from: Scintific Method on March 20, 2013, 08:26:12 PM

Quote

Here we go again. "The earth is round, just like everything in the sky. Except for the things that aren't."

If you think about that for just a couple of seconds, you'll realise that the things that aren't round are all quite a bit smaller than the things that are round

This is incorrect.  I think you'll find that galaxies, according to standard RE theory, are quite large.

The large things that are in galaxies are round, the small ones less so.

I'm not talking about the things in galaxies.  I'm talking about galaxies.  They are huge and they are flat.

Quote

The gravity in a galaxy locally is very weak unlike inside a star.

Your point?  On the whole the gravity in a galaxy is strong enough to hold together billions of stars, or so we are told.  That doesn't seem very weak to me.

And it pulls them together into a flat shape.

The average force of gravity is weaker than in a star or planet. If it were stronger than it would collapse to a sphere. The competing forces, solar winds, gravity from other galaxies, prevents this.

So everything in the universe is compelled by gravity to pull into a sphere... unless it's too small... or unless it's too large... got it.  In other words,

Quote

"The earth is round, just like everything in the sky. Except for the things that aren't."

Not what I said, but take your meaning as you wish.

No, I'd rather we both understand this properly.  How does what I said differ from what you said?

Fair enough, I am going to quote Hippy Sailor, because he said it well:

"Not large or small, but mass. Neutron stars for example are very small as far as stars go, but they have a lot of mass, so they are spherical. A galaxy, taken as a whole, is not very massive. Where there is mass locally, you have star systems, and in the case of a galaxy many localized star systems. Despite these many stars, a galaxy contains far more empty space than anything else.

Mass and size are not the same thing, equal volumes are not necessarily equal in other ways. A gallon of air and a gallon of rocks might occupy the same space, but share few other characteristics. A galaxy at scale is a lot more like air than rocks."

So yes a galaxy's boundaries are large, but proportionate to its size there is not enough mass to make it a sphere, but locally, where there is, you find spheres.

Fair enough, I am going to quote Hippy Sailor, because he said it well:

"Not large or small, but mass. Neutron stars for example are very small as far as stars go, but they have a lot of mass, so they are spherical. A galaxy, taken as a whole, is not very massive. Where there is mass locally, you have star systems, and in the case of a galaxy many localized star systems. Despite these many stars, a galaxy contains far more empty space than anything else.

Mass and size are not the same thing, equal volumes are not necessarily equal in other ways. A gallon of air and a gallon of rocks might occupy the same space, but share few other characteristics. A galaxy at scale is a lot more like air than rocks."

So yes a galaxy's boundaries are large, but proportionate to its size there is not enough mass to make it a sphere, but locally, where there is, you find spheres.

But they are still held together by the force of gravity.  How are galaxies affected so strongly by gravity without being pulled into a sphere?  I mean, I'd get it if you said they're so massless on average that they aren't held together by gravity.  But obviously that's not the case.

The same goes for our solar system.  Why are all the planets on the ecliptic?  Why is it that there aren't any that are orbiting the sun at right angles to our own orbit? 

I'm sorry, but you don't get to say that the existence of gravity means that the Earth must be pulled into a sphere if there are exceptions to the rule.  Would you disagree if I asserted that the known existence of one or two exceptions (lower average mass, much lower total mass) means that there are probably other exceptions that we haven't discovered yet?

Quote from: Rama Set on March 21, 2013, 06:58:25 PM

Fair enough, I am going to quote Hippy Sailor, because he said it well:

"Not large or small, but mass. Neutron stars for example are very small as far as stars go, but they have a lot of mass, so they are spherical. A galaxy, taken as a whole, is not very massive. Where there is mass locally, you have star systems, and in the case of a galaxy many localized star systems. Despite these many stars, a galaxy contains far more empty space than anything else.

Mass and size are not the same thing, equal volumes are not necessarily equal in other ways. A gallon of air and a gallon of rocks might occupy the same space, but share few other characteristics. A galaxy at scale is a lot more like air than rocks."

So yes a galaxy's boundaries are large, but proportionate to its size there is not enough mass to make it a sphere, but locally, where there is, you find spheres.

But they are still held together by the force of gravity.  How are galaxies affected so strongly by gravity without being pulled into a sphere?  I mean, I'd get it if you said they're so massless on average that they aren't held together by gravity.  But obviously that's not the case.

The same goes for our solar system.  Why are all the planets on the ecliptic?  Why is it that there aren't any that are orbiting the sun at right angles to our own orbit? 

I'm sorry, but you don't get to say that the existence of gravity means that the Earth must be pulled into a sphere if there are exceptions to the rule.  Would you disagree if I asserted that the known existence of one or two exceptions (lower average mass, much lower total mass) means that there are probably other exceptions that we haven't discovered yet?

A galaxy is not an exception, I think that it should never have been said that all celestial objects or groups want to form spheres, what should have been said is they want to conserve energy.  For some systems, a sphere is the most energy conservative shape to take, in others a disc.  In regards to galaxies, their gravity is strong enough to keep them associated, but not strong enough to pull them in to spheres.  That being said, a disc is a low energy form to hold and is very good at conserving angular momentum, so this is why it is popular as a galactic shape.  There are other shapes that galaxies take too, like spirals, but these have to do with the pull of neighboring galaxies gravity.  On the scale of stars almost all objects are spherical, likewise on the scale of planets, perhaps it is only on these scales that spheres are common, but they are the most commonly dealt with scale I think it is safe to say.

I hope this clears some of this up.

I hope this clears some of this up.

Not really. You talking in circles merely confirms your mistaken view of the universe and your cultish adherence to the fantasy force of gravity.

Quote from: Rama Set on March 21, 2013, 08:23:01 PM

I hope this clears some of this up.

Not really. You talking in circles merely confirms your mistaken view of the universe and your cultish adherence to the fantasy force of gravity.

I was not talking in circles.  By and large it holds true that most objects we observe on a large scale in space, probably over 99%, are spheres.  Your ad hominem attack does not change anything.

I'm sorry, but you don't get to say that the existence of gravity means that the Earth must be pulled into a sphere if there are exceptions to the rule.  Would you disagree if I asserted that the known existence of one or two exceptions (lower average mass, much lower total mass) means that there are probably other exceptions that we haven't discovered yet?

Yet another attempt at cheap philosophy instead of working with numbers. Your only argument works around the use of words like "must", which implies that matter has free will if you do not understand metaphors.

Just do the actual maths of the problem and you will see that almost any type and shape of solid matter we know will not be static if you have enough of it. For example, take 1x10²² kg (the mass of Pluto) of the best steel on Earth and make a perfectly straight bar, of just a square kilometer of cross section, and calculate the strain on the middle section due to gravity. You will find that it is enough strain to deform the bar, and eventually it will crumble approximately into a sphere.

I have seen someplace in this forum this kind of calculations, and I am not particularly interested in repeating the maths because I do not believe you will even look at it, or maybe even understand it if you try. Find one of the instances in which the calculations have been made in this forum, if you are interested.

Yet another attempt at cheap philosophy instead of working with numbers. Your only argument works around the use of words like "must", which implies that matter has free will if you do not understand metaphors.

No it doesn't.  You have no idea what you're talking about.  Ridiculous accusations like this only serve to muddy the waters and it's all you really know how to do.  Please go away.

Your only argument works around the use of words like "must", which implies that matter has free will if you do not understand metaphors.

If something "must" do something, it means it has free will?  What?

Just do the actual maths of the problem and you will see that almost any type and shape of solid matter we know will not be static if you have enough of it. For example, take 1x10²² kg (the mass of Pluto) of the best steel on Earth and make a perfectly straight bar, of just a square kilometer of cross section, and calculate the strain on the middle section due to gravity. You will find that it is enough strain to deform the bar, and eventually it will crumble approximately into a sphere.

I have seen someplace in this forum this kind of calculations, and I am not particularly interested in repeating the maths because I do not believe you will even look at it, or maybe even understand it if you try. Find one of the instances in which the calculations have been made in this forum, if you are interested.

So, let me get this straight.  You are assuming that the size of Pluto is what you have been taught, then you are speculating that a bar of steel of the same mass (based on the assumed size), would somehow form a ball due to the strain in the center caused by gravity.

Have you ever even taken a physics class?

Have you ever even taken a physics class?

Have you?

Quote from: jroa on March 23, 2013, 04:47:48 AM

Have you ever even taken a physics class?

Have you?

Yes!   

Quote from: Scintific Method on March 23, 2013, 05:28:39 AM

Quote from: jroa on March 23, 2013, 04:47:48 AM

Have you ever even taken a physics class?

Have you?

Yes!   

Wow! So, what would a bar of steel with a mass of 1x10²² and a cross section of 1km² do under the influence of it's own gravity? To help you out, it would be between 1.2x10¹²m and 1.3x10¹²m long (let's just ignore all references to Pluto, and focus on the bar of steel).

It's a bit late at night for me to attempt the maths myself, but I'll certainly be having a look at it when I can.

Quote from: jroa on March 23, 2013, 05:33:34 AM

Quote from: Scintific Method on March 23, 2013, 05:28:39 AM

Quote from: jroa on March 23, 2013, 04:47:48 AM

Have you ever even taken a physics class?

Have you?

Yes!   

Wow! So, what would a bar of steel with a mass of 1x10²² and a cross section of 1km² do under the influence of it's own gravity? To help you out, it would be between 1.2x10¹²m and 1.3x10¹²m long (let's just ignore all references to Pluto, and focus on the bar of steel).

It's a bit late at night for me to attempt the maths myself, but I'll certainly be having a look at it when I can.

I was tempted to plug that into Audtdesk, or another 3D software in order to give you an immediate answer.  But the I realized that you are grasping at straws, an it is actually a little sad.

Quote from: Scintific Method on March 23, 2013, 06:01:42 AM

Quote from: jroa on March 23, 2013, 05:33:34 AM

Quote from: Scintific Method on March 23, 2013, 05:28:39 AM

Quote from: jroa on March 23, 2013, 04:47:48 AM

Have you ever even taken a physics class?

Have you?

Yes!   

Wow! So, what would a bar of steel with a mass of 1x10²² and a cross section of 1km² do under the influence of it's own gravity? To help you out, it would be between 1.2x10¹²m and 1.3x10¹²m long (let's just ignore all references to Pluto, and focus on the bar of steel).

It's a bit late at night for me to attempt the maths myself, but I'll certainly be having a look at it when I can.

I was tempted to plug that into Audtdesk, or another 3D software in order to give you an immediate answer.  But the I realized that you are grasping at straws, an it is actually a little sad.

Not grasping at straws, genuinely interested, which is why I'll be doing the maths myself when I am more awake.